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\begin{center}
{\bf SOLUTIONS FINAL -- PHY 4936 Fall 2011}\\ 
\end{center}
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\bigskip
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\centerline{\bf PROBLEM 1}\bigskip
\noindent
(1) The momentum is
$$ p = \frac{\partial L}{\partial\dot x} = m \dot x$$
\noindent
(2) Eliminating $\dot x$ in favor of $p$ we obtain the Hamiltonian
$$ H = T + V = \frac{p^2}{2m} + \frac{1}{2} k x^2 $$
\noindent
(3) Hamilton's equations are
$$ \frac{\partial H}{\partial p} = \frac{p}{m} = \dot x~~{\rm and}~~ 
   \frac{\partial H}{\partial x} = k x = - \dot p $$
\noindent
(4) Newton's force law follows
$$ k x = - \dot p = -m \ddot x~~{\rm or}~~ m \ddot x = - k x $$
\bigskip
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\centerline{\bf PROBLEM 2}\bigskip
\noindent We consider an infinitesimal rotation by an angle 
$\delta\vec\phi$ (Landau-Lifshitz, \textsection 9, p.18):
\begin{equation} \label{infrot}
  \delta\vec{r}_j\ =\ \delta\vec{\phi}\times\vec{r}_j\,,~~~
  \delta\dot{\vec{r}}_j\ =\ \delta\vec{\phi}\times\dot{\vec{r}}_j\,.
\end{equation}
For example, if this rotation is about the $z$ axis $|\delta\vec{r}_j|= 
|\delta\phi\,r_j\,\sin(\theta)|$ holds. Let us denote the components of 
the rotations by $\delta x^i_j$ and $\delta \dot{x}^i_j$, where 
$i=1,2,3$ labels the coordinates and $j=1,\dots,n$ the particles.
Assuming isotropy of space, the Lagrangian is invariant under an
infinitesimal rotation
$$0\ =\ \sum_j \left\{ \sum_i\frac{\partial L}{\partial x^i_j}\,
        \delta x^i_j + \sum_i\frac{\partial L}{\partial\dot{x}^i_j}\,
        \delta \dot{x}^i_j \right\}\,. $$
Using the definition of the generalized momentum and Euler-Lagrange
equations, this reads
$$ 0\ =\ \sum_j \left\{ \sum_i\dot{p}^i_j\,\delta x^i_j + 
   \sum_i p^i_j\,\delta\dot{x}^i_j \right\}\ =\
   \sum_j \left\{ \dot{\vec{p}}_j\cdot\delta \vec{r}_j + 
   \vec{p}_j\cdot \delta\dot{\vec{r}}_j \right\}\,.$$
Inserting (\ref{infrot}) 
$$ 0\ =\ \sum_j \left\{ \dot{\vec{p}}_j\cdot(\delta\vec{\phi}
   \times\vec{r}_j) + \vec{p}_j\cdot(\delta\vec{\phi}\times
   \dot{\vec{r}}_j) \right\} $$
and we pull out the $\delta\vec{\phi}$, which is the same for all 
particles:
$$ 0\ =\ \sum_j \left\{ \delta\vec{\phi}\cdot(\vec{r}_j\times\dot{p}_j)) 
  + \delta\vec{\phi}\cdot(\dot{\vec{r}}_j\times\vec{p}_j) \right\}\
  =\ \delta\vec{\phi}\ \frac{d~}{dt}\,\sum_j(\vec{r}_j\times\vec{p}_j)$$
$$ \Leftrightarrow\ \sum_j(\vec{r}_j\times\vec{p}_j)\ =\ \vec{L}\
  =\ {\rm Constant}\,. $$
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\centerline{\bf PROBLEM 3: See Homework, Problem~19.} \bigskip

\centerline{\bf PROBLEM 4: Compare Homework, Problem~29.} \bigskip

\noindent 1. Given a coordinate system $(x',y')$ which rotates with the 
disk, the location of the CM is $\bar{x}_{CM}=0$ and
\begin{eqnarray*}
  \bar{y}_{CM}&=& \frac{\rho}{M}\left\{\int_0^R drr\int_0^{\pi}d\theta 
     r\sin\theta+2\int_0^R dr r\int_{\pi}^{2\pi}d\theta r 
     \sin\theta\right\} \\
  &=&\frac{\rho}{M}\left\{\frac{R^3}{3}2-\frac{R^3}{3}4\right\}
   = -\frac{2}{3}\frac{\rho R^3}{M}=-\frac{4R}{9\pi}~~{\rm as}~~
   M\ =\ \frac{3\pi\,\rho\,R^2}{2}\,.
\end{eqnarray*}
\bigskip

\noindent 2. The relationship between the lab coordinates and the CM
coordinates is
\begin{eqnarray} \nonumber
  x_{CM}&=&R\theta-|\bar{y}_{CM}|\sin\theta\ 
         =\ R\theta-(4R/9\pi)\,\sin\theta\,,\\ \nonumber
  y_{CM}&=&R-|\bar{y}_{CM}|\cos\theta~\
         =\ R-(4R/9\pi)\,\cos\theta\,.
\end{eqnarray} 
\bigskip

\noindent 3.  To find $I_3^{CM}$ we calculate first $I_3^0$ with respect 
to the center of disk and use the parallel axis theorem, (32.12) of 
Landau and Lifshitz, 
\begin{eqnarray*}
  I_3\Bigl\vert_0&=&\rho\int_0^R dr r\int_0^{\pi}d\theta (x^2+y^2)+2
  \rho\int _0^R dr r\int _{\pi}^{2\pi}d\theta (x^2+y^2)\\
  &=&\rho\frac{R^4}{4}\pi+2\rho\frac{R^4}{4}\pi=\rho R^4\frac{3}{4}\pi
  =\frac{1}{2}MR^2\\
  I_3\Bigl\vert_{CM}&=&I_3\bigl\vert_{0}-M\bar{y}_{CM}^2
  =\frac{1}{2}MR^2-M\frac{16}{81}\frac{R^2}{\pi^2}\
  =\ \frac{1}{2}MR^2\left[ 1-\frac{32}{81\pi^2}\right].
\end{eqnarray*}
\bigskip

\noindent 4. The Lagrangian of the disk is 
$$ L\ =\ \frac{1}{2}M(\dot{x}_{CM}^2+\dot{y}_{CM}^2)+
         \frac{1}{2}I_3 \dot\theta^2 - M\,g\,y_{CM}\,.  $$ 

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